Circle geometry

1. Parts of circles

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Tangent
Chord (non-diameter)
Diameter
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Tangent F
Chord (non-diameter) B
Diameter D
NOTE: a diameter is also a chord. However, this question asks you to identify a chord which is not a diameter, so the correct answer for Chord must be "B" and cannot be "D."
STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle.
Any straight line segment from the centre of the circle to a point on the circumference.
A straight line segment joining the ends of an arc.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle. Diameter
Any straight line segment from the centre of the circle to a point on the circumference. Radius
A straight line segment joining the ends of an arc. Chord

Submit your answer as: andand

Vocabulary: subtends

The figure below shows a circle. There are three points labelled on the circumference, K,J, and H. There are also two angles, labelled z and x.

Which angle(s) is subtended by chord JK?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to chord JK.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about chord JK: this chord is highlighted below in dark red. It stretches from point J to point K.

The shaded section shows the chord and the angle which spreads out to meet that chord.

Therefore the correct answer is that chord JK subtends angle z.


Submit your answer as:

Exercises

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Tangent
Radius
Chord (non-diameter)
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Tangent F
Radius C
Chord (non-diameter) B
NOTE: a diameter is also a chord. However, this question asks you to identify a chord which is not a diameter, so the correct answer for Chord must be "B" and cannot be "D."
STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Radius
Segment
Arc
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Radius C
Segment E
Arc A

STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

Circle geometry terminology

The following diagram is colour coded. In this diagram you see the different parts of the circle labelled from A-F, each label is the same colour as the part of the circle it describes.

Select the label from Column B that best fits the word in Column A:

Answer:
Column A Column B
Tangent
Radius
Arc
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]
Column A Column B
Tangent F
Radius C
Arc A

STEP: <no title>
[−0 points ⇒ 0 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


Submit your answer as: andand

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle.
A part of the circumference of a circle.
A line that makes contact with a circle at one point on the circle.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle. Diameter
A part of the circumference of a circle. Arc
A line that makes contact with a circle at one point on the circle. Tangent

Submit your answer as: andand

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
Any straight line segment from the centre of the circle to a point on the circumference.
A line that makes contact with a circle at one point on the circle.
A part of the circumference of a circle.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
Any straight line segment from the centre of the circle to a point on the circumference. Radius
A line that makes contact with a circle at one point on the circle. Tangent
A part of the circumference of a circle. Arc

Submit your answer as: andand

Circle geometry terminology

Select a word from Column B that best fits the description in Column A:

Answer:
Column A Column B
Any straight line segment from the centre of the circle to a point on the circumference.
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle.
A part of the circumference of a circle.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The following diagram is colour coded. In this diagram you see the different terms used when referring to circles. Each term is the same colour as the part of the circle it describes.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]
Column A Column B
Any straight line segment from the centre of the circle to a point on the circumference. Radius
A special chord that stretches from one point on the circumference to another point on the circumference and that passes through the centre of a circle. Diameter
A part of the circumference of a circle. Arc

Submit your answer as: andand

Vocabulary: subtends

The figure below shows a circle. There are two points labelled on the circumference, K, and J. There are also two angles, labelled a and c.

Arc JK subtends which angle(s)?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to arc JK.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about arc JK: this arc is highlighted below in dark red. It stretches from point J to point K.

There are two shaded sections: one in blue and one with bars which are light blue. The arc subtends both of these angles - both of the angles spread out to meet the arc.

Therefore the correct answer is that arc JK subtends both of the angles in the figure: angle a and angle c.


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Vocabulary: subtends

The figure below shows a circle. There are two points labelled on the circumference, W, and X. There are also two angles, labelled n and e.

Which angle(s) is subtended by arc WX?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to arc WX.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about arc WX: this arc is highlighted below in dark red. It stretches from point W to point X.

There are two shaded sections: one in blue and one with bars which are light blue. The arc subtends both of these angles - both of the angles spread out to meet the arc.

Therefore the correct answer is that arc WX subtends both of the angles in the figure: angle e and angle n.


Submit your answer as:

Vocabulary: subtends

The figure below shows a circle. There are three points labelled on the circumference, A,B, and C. There are also two angles, labelled c and a.

Which angle(s) is subtended by chord AB?

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

'Subtends' means 'maps a path to' or 'matches the spread of.' We need to find the angle (or angles) which maps out, or spreads out, to chord AB.


STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Think of light coming through a small window: the light crosses the room and makes a bright patch of light somewhere in the room, maybe on the floor or on a wall. An arc or chord subtending an angle is kind of like that.

The window is like the angle, where the light starts, and the patch of light is like the arc/chord. If an arc or chord subtends an angle, then you can draw lines directly from the vertex of the angle to the endpoints of the arc/chord. (Like the light crossing the room from the window to the patch of light!)

The question asks us about chord AB: this chord is highlighted below in dark red. It stretches from point A to point B.

The shaded section shows the chord and the angle which spreads out to meet that chord.

Therefore the correct answer is that chord AB subtends angle c.


Submit your answer as:

2. Chord theorems

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. S and T are points on AC and AB respectively such that OSAC and OTAB. AB=136 and AC=168.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AT.

    Answer:

    AT=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AT. We know that AB has a length of 136. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OT is perpendicular to chord AB. Therefore OT bisects AB. And that means AT is half as long as AB. We can use that fact to find the length of AT.

    AT=AB2=1362=68

    AT has a length of 68.


    Submit your answer as:
  2. If OS=1351OT, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOSA, side SA must be half as long as AC. This is for the same reason we used in Question 1. So SA=84, as labelled below. We are using d for the length of OT.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOSA:

    c2=a2+b2r2=OS2+SA2r2=(1351d)2+(84)2

    We can write a similar equation for ΔOTA.

    c2=a2+b2r2=OT2+TA2r2=(d)2+(68)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+682=(1351d)2+842
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OT).

    d2+682=(1351)2d2+842d2+4624=1692601d2+7056d21692601d2=7056462424322601d2=2432d2=2601d=±51

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(68)2r2=(51)2+4624r2=2601+4624=7225r=±85

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 85.


    Submit your answer as:

Using the perpendicular bisector theorem

In the circle with centre O, segment CJ¯=12. OC¯ is perpendicular to GJ¯ at Point C. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    GJ¯ is because .

    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, GJ¯ is the chord, and it is bisected at point C. In other words, C is the midpoint of GJ¯. We know the chord is bisected because it is perpendicular to segment OC¯.

    On the diagram below the segments GC¯ and CJ¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: GJ¯ is "twice as long as CJ¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of GJ¯ in the diagram.

    Answer: GJ¯=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is larger than the length given. Therefore, we must multiply by two to get the length of the chord.

    larger=2×shorterGJ¯=2(CJ¯)=2(12)=24

    The length of GJ¯ is 24.


    Submit your answer as:

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord HR¯=8 and OF¯=4. Also, OF¯HR¯ at point F. Determine the length of OR¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OR¯ is .
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord HR¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of HR¯=8
  • length of OF¯=4
  • OF¯ is perpendicular to HR¯

and we want:

  • the length of the side OR¯ which is the radius of the circle

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord HR¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point F. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that HF¯ and FR¯ are half the size of HR¯, which is 8.

HF¯=FR¯=HR¯2(line from centre bisects chord)=(8)2=4

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OR¯ is the radius . This is also the line segment you want!.

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OF¯ is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2(OR¯)2=(OF¯)2+(FR¯)2=(4)2+(4)2OR¯=±16+16=±32

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OR¯=5,66.


Submit your answer as:

Using circle theorems to relate angles

In the circle with midpoint O the arc CA subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is CA and the angles x and y are both subtended by it. Therefore, the angles must be the same size: angle y in terms of x is y=x.

The reason for this theorem is: s subtended by same arc CA.


Submit your answer as: and

Exercises

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. S and T are points on AC and AB respectively such that OSAC and OTAB. AB=40 and AC=48.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AT.

    Answer:

    AT=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AT. We know that AB has a length of 40. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OT is perpendicular to chord AB. Therefore OT bisects AB. And that means AT is half as long as AB. We can use that fact to find the length of AT.

    AT=AB2=402=20

    AT has a length of 20.


    Submit your answer as:
  2. If OS=715OT, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOSA, side SA must be half as long as AC. This is for the same reason we used in Question 1. So SA=24, as labelled below. We are using d for the length of OT.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOSA:

    c2=a2+b2r2=OS2+SA2r2=(715d)2+(24)2

    We can write a similar equation for ΔOTA.

    c2=a2+b2r2=OT2+TA2r2=(d)2+(20)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+202=(715d)2+242
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OT).

    d2+202=(715)2d2+242d2+400=49225d2+576d249225d2=576400176225d2=176d2=225d=±15

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(20)2r2=(15)2+400r2=225+400=625r=±25

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 25.


    Submit your answer as:

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. S and T are points on AC and AB respectively such that OSAC and OTAB. AB=40 and AC=48.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AT.

    Answer:

    AT=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AT. We know that AB has a length of 40. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OT is perpendicular to chord AB. Therefore OT bisects AB. And that means AT is half as long as AB. We can use that fact to find the length of AT.

    AT=AB2=402=20

    AT has a length of 20.


    Submit your answer as:
  2. If OS=715OT, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOSA, side SA must be half as long as AC. This is for the same reason we used in Question 1. So SA=24, as labelled below. We are using d for the length of OT.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOSA:

    c2=a2+b2r2=OS2+SA2r2=(715d)2+(24)2

    We can write a similar equation for ΔOTA.

    c2=a2+b2r2=OT2+TA2r2=(d)2+(20)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+202=(715d)2+242
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OT).

    d2+202=(715)2d2+242d2+400=49225d2+576d249225d2=576400176225d2=176d2=225d=±15

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(20)2r2=(15)2+400r2=225+400=625r=±25

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 25.


    Submit your answer as:

Using circle geometry to find the radius

Adapted from DBE Nov 2016 Grade 12, P2, Q8.2
Maths formulas

A,B, and C are points on the circle having centre O. X and Y are points on AC and AB respectively such that OXAC and OYAB. AB=136 and AC=154.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Calculate AX.

    Answer:

    AX=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the theorem about the perpendicular lines from the centre of a circle.


    STEP: Apply the perpendicular line from the centre of a circle theorem
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the length of AX. We know that AC has a length of 154. To find the answer, we can use the theorem about perpendicular lines from the centre of a circle. A line through the centre of a circle which is perpendicular to a chord bisects that chord.

    The segment OX is perpendicular to chord AC. Therefore OX bisects AC. And that means AX is half as long as AC. We can use that fact to find the length of AX.

    AX=AC2=1542=77

    AX has a length of 77.


    Submit your answer as:
  2. If OX=1217OY, calculate the radius OA of the circle.

    INSTRUCTION: Round your answer to two decimal places, if necessary.
    Answer:

    OA=

    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Use the right-angled triangles in the diagram to write two equations. Use the theorem of Pythagoras.


    STEP: Label the diagram with known information
    [−1 point ⇒ 4 / 5 points left]

    We need to find the radius of the circle (the length of OA). Let's start by labelling the information we know about the figure, and the radius that we want to find. In ΔOYA, side YA must be half as long as AB. This is for the same reason we used in Question 1. So YA=68, as labelled below. We are using d for the length of OY.


    STEP: Write an equation for each of the right-angled triangles
    [−1 point ⇒ 3 / 5 points left]

    There are two right-angled triangles, and they share a side. In fact, the side they share is the radius that we are looking for. We can write an equation for each triangle using the theorem of Pythagoras. In ΔOXA:

    c2=a2+b2r2=OX2+XA2r2=(1217d)2+(77)2

    We can write a similar equation for ΔOYA.

    c2=a2+b2r2=OY2+YA2r2=(d)2+(68)2

    STEP: Combine the equations and solve for d
    [−2 points ⇒ 1 / 5 points left]

    Now we have two equations that are both equal to r2. So they must be equal to each other, and we can write:

    d2+682=(1217d)2+772
    NOTE: We are solving simultaneous equations. The equation above comes from combining the equations by substitution.

    Now solve the equation for d (which is the length of OY).

    d2+682=(1217)2d2+772d2+4624=144289d2+5929d2144289d2=59294624145289d2=1305d2=2601d=±51

    The result we want is a distance, so we can throw away the negative answer.


    STEP: Calculate the length of the radius
    [−1 point ⇒ 0 / 5 points left]

    We can use the result above to calculate the length of the radius. We can use either of the equations above. We will use the second equation (because the first equation contains an extra fraction that we can avoid).

    r2=(d)2+(68)2r2=(51)2+4624r2=2601+4624=7225r=±85

    Again, we throw away the negative answer because we are calculating a distance. The length of radius OA is 85.


    Submit your answer as:

Using the perpendicular bisector theorem

In the circle with centre O, segment GP¯=11. OG¯ is perpendicular to AP¯ at Point G. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    AP¯ is because .

    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, AP¯ is the chord, and it is bisected at point G. In other words, G is the midpoint of AP¯. We know the chord is bisected because it is perpendicular to segment OG¯.

    On the diagram below the segments AG¯ and GP¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: AP¯ is "twice as long as GP¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of AP¯ in the diagram.

    Answer: AP¯=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is bigger than the length given. Therefore, we must multiply by two to get the length of the chord.

    bigger=2×shorterAP¯=2(GP¯)=2(11)=22

    The length of AP¯ is 22.


    Submit your answer as:

Using the perpendicular bisector theorem

In the circle with centre O, segment KC¯=18. OC¯ is perpendicular to KH¯ at Point C. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    KH¯ is because .

    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, KH¯ is the chord, and it is bisected at point C. In other words, C is the midpoint of KH¯. We know the chord is bisected because it is perpendicular to segment OC¯.

    On the diagram below the segments KC¯ and CH¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: KH¯ is "twice as long as KC¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of KH¯ in the diagram.

    Answer: KH¯=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is bigger than the length given. Therefore, we must multiply by two to get the length of the chord.

    bigger=2×shorterKH¯=2(KC¯)=2(18)=36

    The length of KH¯ is 36.


    Submit your answer as:

Using the perpendicular bisector theorem

In the circle with centre O, segment DF¯=8. OQ¯ is perpendicular to DF¯ at Point Q. This is shown in the figure below, which is drawn to scale.

  1. Complete the sentence below by choosing the correct options.

    Answer:

    QF¯ is because .

    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    We have a circle theorem which says the following: if a line passing through the centre of a circle is perpendicular to a chord in the circle, then the line bisects the chord. 'Bisect' means 'cuts in half' or 'cuts in two equal pieces.'


    STEP: Use the perpendicular bisector theorem
    [−2 points ⇒ 0 / 2 points left]

    One of the circle theorems tells us that if a line from the centre of the circle is perpendicular to any chord in the circle, then the line from the center bisects the chord. 'Bisect' means 'cut in two equal pieces' or 'cuts in half.'

    In this figure, DF¯ is the chord, and it is bisected at point Q. In other words, Q is the midpoint of DF¯. We know the chord is bisected because it is perpendicular to segment OQ¯.

    On the diagram below the segments DQ¯ and QF¯ are shaded in blue and red. You can see that they look about the same size (the diagram is to scale so it is ok to trust your eyes a bit).

    Based on this, we can get the answers: QF¯ is "half the length of DF¯" because " line from centre bisects chord" (a line drawn from the centre of a circle, perpendicular to a chord, bisects the chord).


    Submit your answer as: and
  2. Now find the length of QF¯ in the diagram.

    Answer: QF¯=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the segment you want to find is the smaller one, then you need to divide by two. If the segment you want is the larger one, you need to multiply by two!


    STEP: Use the answer we found in Question 1 to calculate the answer
    [−1 point ⇒ 0 / 1 points left]

    The length that we want is smaller than the length given. Therefore, we must divide by two to get the length of the smaller segment (this is the same as multiplying by a half).

    smaller=12×largerQF¯=12(DF¯)=12(8)=4

    The length of QF¯ is 4.


    Submit your answer as:

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord QD¯=6 and OR¯=3. Also, OR¯QD¯ at point R. Determine the length of OQ¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OQ¯ is .
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord QD¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of QD¯=6
  • length of OR¯=3
  • OR¯ is perpendicular to QD¯

and we want:

  • the length of the side OQ¯ which is the radius of the circle

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord QD¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point R. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that QR¯ and RD¯ are half the size of QD¯, which is 6.

QR¯=RD¯=QD¯2(line from centre bisects chord)=(6)2=3

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OQ¯ is the radius . This is also the line segment you want!.

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OR¯ is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2(OQ¯)2=(OR¯)2+(RQ¯)2=(3)2+(3)2OQ¯=±9+9 =±18

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OQ¯=4,24.


Submit your answer as:

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord BJ¯=8 and OK¯=6. Also, OK¯BJ¯ at point P. Find the length of OP¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OP¯ is .
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord BJ¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of BJ¯=8
  • length of OK¯=6 which is the radius of the circle
  • OK¯ is perpendicular to BJ¯

and we want:

  • the length of the side OP¯

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord BJ¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point P. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that BP¯ and PJ¯ are half the size of BJ¯, which is 8.

BP¯=PJ¯=BJ¯2(line from centre bisects chord)=(8)2=4

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OK¯ is the radius of the circle; and more importantly, it is equal to any other radius of the circle, e.g. OK¯OB¯.

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OP¯, the line segment you want to find, is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2OB¯2=OP¯2+PB¯2(6)2=OP¯2+(4)2OP¯=±(36)(16)=±20

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

Finally, remember that the question asked you to, 'round your answer to two decimal places (if necessary).'

The final answer is: OP¯=4,47.


Submit your answer as:

Calculations using the 'perpendicular bisector' theorem

In the circle O you are given the length of the chord RG¯=8 and OQ¯=3. Also, OH¯RG¯ at point Q. Compute the length of OH¯.

INSTRUCTION: Round your answer to two decimal places (if necessary).
Answer: The length of OH¯ is .
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Notice there is a right angle in this diagram at chord RG¯. This means you can use the perpendicular bisector theorem.


STEP: Use the perpendicular bisector theorem
[−1 point ⇒ 3 / 4 points left]

Let's organise the information before we get started. We know:

  • the length of RG¯=8
  • length of OQ¯=3
  • OH¯ is perpendicular to RG¯

and we want:

  • the length of the side OH¯ which is the radius of the circle

There are several ways to tackle this problem! Let's look in the figure to see if there is a theorem about circles which can be helpful.

We have the length of the chord RG¯, and there is a line from the centre of the circle O perpendicular to the chord, meeting the chord at point Q. What a surprise... no, not really: we can use the theorem about the perpendicular bisectors through the centre of a circle. In this case, it tells us that RQ¯ and QG¯ are half the size of RG¯, which is 8.

RQ¯=QG¯=RG¯2(line from centre bisects chord)=(8)2=4

STEP: Set up the information you have about a right-angled triangle
[−1 point ⇒ 2 / 4 points left]

It is also important to notice that OH¯ is the radius of the circle; and more importantly, it is equal to any other radius of the circle, e.g. OH¯OR¯. This is also the line segment you want!.

The key to opening this solution is to see that there is a right-angled triangle in the figure. Can you see it? Line segment OQ¯ is one of the sides of this triangle.

Now we have three sides of a right-angled triangle.


STEP: Use the Theorem of Pythagoras
[−2 points ⇒ 0 / 4 points left]

We want one of the sides of a right triangle: it is Pythagoras time! (We will not get the answer using sine, cosine or tangent because we don't know either of the acute angles in the triangle.)

c2=a2+b2OR¯2=OQ¯2+QR¯2OR¯2=(3)2+(4)2OR¯=±(9)+(16)=±25=±5

Cool! We can throw out the negative sign, because this number is a distance - it has to be positive!

But wait! The question asked for the length of OH¯ and we just found the length of OR¯. Was this a waste of time? No! You have found the radius of the circle: OR¯=25; and OH¯ is also a radius of the circle, so it has the same length! Therefore, OH¯=25.

The question says that you should round the answer to two decimal places; but in this case the answer works out to be an integer, so there is no surd or decimals to deal with.

The final answer is: OH¯=5.


Submit your answer as:

Using circle theorems to relate angles

In the circle with midpoint M the arc EK subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

A useful circle theorem for this question is the following: the angle subtended by an arc at the centre of a circle is twice the size of the angle subtended by the same arc at the circumference of the circle. Here the arc is EK; the angle subtended by the arc at the centre of a circle is x; and the angle subtended by the same arc at the circumference of the circle is y. Therefore, x is twice the size of y or: x=2y. Writing angle y in terms of x, we get y=12x.

The reason for this theorem is:  at centre =2 at circum.


Submit your answer as: and

Using circle theorems to relate angles

In the circle with midpoint M the arc KJ subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

We recall the circle theorem which states that angles subtended by arcs of equal length, or by the same arc, are equal. Here the arc is KJ and the angles x and y are both subtended by it. Therefore, the angles must be the same size: angle y in terms of x is y=x.

The reason for this theorem is: s subtended by same arc KJ.


Submit your answer as: and

Using circle theorems to relate angles

In the circle with midpoint M the arc ED subtends the angles x and y. What is angle y in terms of x? Please choose a reason for your answer from the options given.

Answer: y= because:
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The words "in terms of x" mean that the answer will include x - it will not be a number alone. For example, 2x3x and 5x1 are "in terms of x."
STEP: Determine which circle theorem applies to this question, and then use it
[−2 points ⇒ 0 / 2 points left]

A useful circle theorem for this question is the following: the angle subtended by an arc at the centre of a circle is twice the size of the angle subtended by the same arc at the circumference of the circle. Here the arc is ED; the angle subtended by the arc at the centre of a circle is x; and the angle subtended by the same arc at the circumference of the circle is y. Therefore, x is twice the size of y or: x=2y. Writing angle y in terms of x, we get y=12x.

The reason for this theorem is:  at centre =2 at circum.


Submit your answer as: and

3. Cyclic quadrilateral theorems

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: A,B,C,D, and E. There is also one angle given: EB^C=60°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point A because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'BCDE' or 'BEDC' or 'CBED' etc. However, you may not write 'bcde' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    BCDECDEBDEBCEBCDEDCBBEDCCBEDDCBE

    Submit your answer as:
  2. What are the measures of the angles CD^E and DE^B in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of CD^E= °
    2. The measure of DE^B=°
    numeric
    string
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: B^ and D^ are opposite each other; C^ and E^ are opposite from each other.

    We know that B^ and D^ have a sum of 180°:

    B^+D^=180°(60°)+D^=180°D^=180°60°D^=120°

    The same relationship is true for C^ and E^ (they are supplementary). However, we cannot find the answer for E^ because we don't know the value of C^. (All we can say is that together they make 180°.) So we cannot determine the measure for this angle.

    The answers are:

    1. The answer for CD^E is 120°.
    2. The answer for DE^B is no solution.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral ABCD is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord at the centre of the circle is twice the angle subtended by the same chord the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that A+C=180°.

    Answer:
    Statement Reason
    at centre = 2 at circum.
    O2=2C at centre = 2 at circum.
    s in a rev.
    A+C=180°
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2A at centre = 2 at circum.
    O2=2C at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2A+2C=360°
    A+C=180°

    Submit your answer as: andand

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 5 points: P,Q,R,S, and T.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point R because it is not on the circle.

    The figure shows 3 points on the circle, which are circled in red. (Point Q is inside of the circle, not on the circumference!)

    There is no cyclic shape in this figure.

    The correct answer from the list is: there is no cyclic quadrilateral.


    Submit your answer as:
  2. Which of the following statements is true about cyclic quadrilaterals?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    One important fact about cyclic quadrilaterals tells us about the size of opposite angles.
    STEP: Use the facts about angles in cyclic quadrilaterals
    [−1 point ⇒ 0 / 1 points left]

    You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

    The correct answer is opposite angles have a sum of 180 degrees.


    Submit your answer as:

Exercises

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: P,Q,R,S, and T. There are also two angles given: R^=72° and S^=109°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point P because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'RQTS' or 'QRST' or 'TSRQ' etc. However, you may not write 'rqts' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    QRSTRSTQSTQRTQRSTSRQQTSRRQTSSRQT

    Submit your answer as:
  2. What are the measures of the angles ST^Q and TQ^R in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of ST^Q= °
    2. The measure of TQ^R=°
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: Q^ and S^ are opposite each other; R^ and T^ are opposite from each other.

    We know that R^ and T^ have a sum of 180°:

    R^+T^=180°(72°)+T^=180°T^=180°72°T^=108°

    It is the same story over again for angles S^ and Q^Q^=180109°=71°.

    The answers are:

    1. The answer for ST^Q is 108°.
    2. The answer for TQ^R is 71°.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: A,B,C,D, and E. There are also two angles given: E^=100° and A^=61°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point C because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'ABDE' or 'AEDB' or 'DEAB' etc. However, you may not write 'abde' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    ABDEBDEADEABEABDEDBAAEDBBAEDDBAE

    Submit your answer as:
  2. What are the measures of the angles AB^D and BD^E in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of AB^D= °
    2. The measure of BD^E=°
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: A^ and D^ are opposite each other; B^ and E^ are opposite from each other.

    We know that E^ and B^ have a sum of 180°:

    E^+B^=180°(100°)+B^=180°B^=180°100°B^=80°

    It is the same story over again for angles A^ and D^D^=18061°=119°.

    The answers are:

    1. The answer for AB^D is 80°.
    2. The answer for BD^E is 119°.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

Opposite angles of a cyclic quadrilateral: Theorem

The figure below shows a circle with centre O and 5 points: G,H,I,J, and K. There are also two angles given: J^=131° and K^=96°.

  1. Name the cyclic quadrilateral in this figure.

    INSTRUCTION: If there is no cyclic quadrilaterial in the figure, write none.
    Answer:

    The cyclic quadrilateral is .

    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To name a quadrilateral, you must name all four of its vertices: pick any vertex of the quadrilateral and then go around the perimeter of the quadrilateral and add each vertex as you come to it.


    STEP: Identify the quadrilateral and find its name
    [−1 point ⇒ 0 / 1 points left]

    The figure shows the four points which lie on the circle. They are the vertices of the quadrilateral (they are circled in red below). Naming the cyclic quadrilateral means that we must name all four of those vertices as we go around the perimeter of the quadrilateral. (We can ignore point I because it is not on the circle.)

    NOTE: There are many acceptable ways to name the quadrilateral: you can write 'HJKG' or 'KGHJ' or 'JKGH' etc. However, you may not write 'hjkg' because the points are named with capital letters and you cannot change them.

    The answer can be any list of the four points on the circumference of the circle, as long as they follow the order of the points around the circle. Any of these answers is correct:

    GHJKHJKGJKGHKGHJKJHGGKJHHGKJJHGK

    Submit your answer as:
  2. What are the measures of the angles KG^H and GH^J in the quadrilateral?

    INSTRUCTION: If either answer is impossible to determine, type no solution.
    Answer:
    1. The measure of KG^H= °
    2. The measure of GH^J=°
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Remember that we know a theorem about the opposite angles of a cyclic quadrilateral: they are supplementary.


    STEP: Use the fact that opposite angles of cyclic quadrilaterals are supplementary, if possible
    [−2 points ⇒ 0 / 2 points left]

    We must use this fact about cyclic quadrilaterals: their opposite angles are supplementary (have a sum of 180 degrees). We can use that fact to answer the question.

    The quadrilateral has two pairs of opposite angles: G^ and J^ are opposite each other; H^ and K^ are opposite from each other.

    We know that J^ and G^ have a sum of 180°:

    J^+G^=180°(131°)+G^=180°G^=180°131°G^=49°

    It is the same story over again for angles K^ and H^H^=18096°=84°.

    The answers are:

    1. The answer for KG^H is 49°.
    2. The answer for GH^J is 84°.

    Submit your answer as: and
  3. Do you solve Question 2 above using a Theorem or its converse?

    Answer: We use the .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theorems state a connection between two things, and the converse states the same connection 'backwards.' Does the work you did above use the following connection straight through, or does it go backwards: 'Opposite angles in a cyclic quadrilateral are supplementary.'


    STEP: Compare the theorem and the converse to Question 2
    [−1 point ⇒ 0 / 1 points left]

    The theorem about the angles in cyclic quadrilaterals states the following connection:

    • Theorem: If a quadrilateral is cyclic the opposite angles must be supplementary.

    The converse states the same connection but in 'reverse:'

    • Converse: If the opposite angles of a quadrilateral are supplementary the quadrilateral must be cyclic.

    Solving the questions above uses the theorem: because we know the quadrilateral is cyclic, then we also know that the angles are supplementary.

    We solve Question 2 using a theorem.


    Submit your answer as:

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral WXYZ is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord a circle is twice the angle subtended by the same chord at the circumference of the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that W+Y=180°.

    Answer:
    Statement Reason
    at centre = 2 at circum.
    O2=2Y at centre = 2 at circum.
    s in a rev.
    W+Y=180°
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2W at centre = 2 at circum.
    O2=2Y at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2W+2Y=360°
    W+Y=180°

    Submit your answer as: andand

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral KLMN is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord a circle is twice the angle subtended by the same chord at the circumference of the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that K+M=180°.

    Answer:
    Statement Reason
    O1=2K at centre = 2 at circum.
    at centre = 2 at circum.
    s in a rev.
    K+M=180°
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2K at centre = 2 at circum.
    O2=2M at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2K+2M=360°
    K+M=180°

    Submit your answer as: andand

Circle theorems

Adapted from DBE Nov 2015 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, cyclic quadrilateral ABCD is drawn in the circle with centre O.

  1. Complete the following statement:

    Answer:

    The angle subtended by a chord a circle is twice the angle subtended by the same chord at the circumference of the circle.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise circle geometry in the Everything Maths textbook.


    STEP: Complete the statement
    [−1 point ⇒ 0 / 1 points left]
    The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference of the circle.

    Submit your answer as:
  2. Complete the table below to prove that A+C=180°.

    Answer:
    Statement Reason
    at centre = 2 at circum.
    O2=2C at centre = 2 at circum.
    s in a rev.
    A+C=180°
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the statement from Question 1 to start the proof.


    STEP: Complete the proof
    [−3 points ⇒ 0 / 3 points left]

    The table below shows the completed proof, that opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

    Statement Reason
    O1=2A at centre = 2 at circum.
    O2=2C at centre = 2 at circum.
    O1+O2=360° s in a rev.
    2A+2C=360°
    A+C=180°

    Submit your answer as: andand

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 4 points: P,Q,R, and S.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point R because it is not on the circle.

    The figure shows 2 points on the circle, which are circled in red. (Point Q is inside of the circle, not on the circumference!)

    There is no cyclic shape in this figure.

    The correct answer from the list is: there is no cyclic quadrilateral.


    Submit your answer as:
  2. Which of the following statements is true about cyclic quadrilaterals?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    One important fact about cyclic quadrilaterals tells us about the size of opposite angles.
    STEP: Use the facts about angles in cyclic quadrilaterals
    [−1 point ⇒ 0 / 1 points left]

    You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

    The correct answer is opposite angles have a sum of 180 degrees.


    Submit your answer as:

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 5 points: G,H,I,J, and K.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point G because it is not on the circle.

    The figure shows 4 points on the circle, which are circled in red.

    Because the shape has four sides and the vertices are all on the edge of the circle, the figure HIJK is a cyclic quadrilateral. The cyclic quadrilateral is shown shaded in light blue.

    The correct answer from the list is: HIJK.


    Submit your answer as:
  2. In the cyclic quadrilateral, which angle is opposite to H^?

    Answer: The angle is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Start at point H and drag your finger across the quadrilateral to see which point is on the other side. When you do this, your finger will go between the other two points which make the quadrilateral.
    STEP: Identify pairs of opposite angles
    [−1 point ⇒ 0 / 1 points left]

    The quadrilateral has two pairs of opposite angles: H^ and J^ are opposite each other; I^ and K^ are opposite from each other.

    The correct answer is J.


    Submit your answer as:

What do cyclic quadrilaterals look like?

The figure below shows a circle with centre O and 4 points: A,B,C, and D.

Answer the two questions below about the figure.

  1. Identify a cyclic quadrilateral in the figure from the list below.

    Answer: The cyclic quadrilateral is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    'Cyclic' means 'on a circle' or 'following a circular path.' A quadrilateral is a shape with four sides. Together, the words 'cyclic quadrilateral' mean a shape with four sides and with vertices sitting on a circle.


    STEP: Look for four points on the circumference of the circle
    [−1 point ⇒ 0 / 1 points left]

    Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle (concyclic). We need to focus on the points which are on the circumference of the circle: for example, we can ignore point B because it is not on the circle.

    The figure shows 3 points on the circle, which are circled in red.

    There is a cyclic triangle in this figure, but not a cyclic quadrilateral.

    The correct answer from the list is: there is no cyclic quadrilateral.


    Submit your answer as:
  2. Which of the following statements is true about cyclic quadrilaterals?

    Answer: The true statement is: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    One important fact about cyclic quadrilaterals tells us about the size of opposite angles.
    STEP: Use the facts about angles in cyclic quadrilaterals
    [−1 point ⇒ 0 / 1 points left]

    You need to know the following about cyclic quadrilaterals: (a) the vertices of the quadrilateral must sit on the edge (circumference) of a circle, and (b) opposite angles of a cyclic quadrilateral are supplementary (they have a sum of 180 degrees).

    The correct answer is opposite angles have a sum of 180 degrees.


    Submit your answer as:

4. Tangent theorems

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line UX¯ which touches the circle at point V. Three points lie on the circle, V,Y and Z, and five angles are labelled: 1,2,3, etc.

Idenitify an angle which is congruent to angle 1. (If there is no angle congruent to 1, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (UX¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, V: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle 1

In the figure below, the relevant information is shown in red. It is very helpful to see the figure below as a tangent line and a triangle - the tangent-chord theorem will always include that triangle!

Notice that one of the angles is inside the triangle, while the other angle is outside of the triangle.

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle 2 to the red chord... then from the red chord follow the red arrow across the circle to angle 1 which subtends the chord (angle 1 is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles we just connected are congruent (the same size): angle 1 angle 2. By looking at these two angles, you can see that they do look like they are about the same size; the tangent-chord theorem says that they must be the same size!


Submit your answer as:

Applying theorems from Euclidean geometry

The figure below shows a cyclic quadrilateral with vertices at the points A,B,C, and D. There are also two angles given: B^=x35° and C^=x4+65°.

  1. Determine D^ in terms of x. If the answer cannot be determined, write 'No solution'.

    Answer: measure of D^=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    When you read "in terms of x," it means that the answer will be a maths expression which includes the variable x.
    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    There is a lot of information in the question about the diagram, but the diagram itself does not show that information; start by labelling the things that you know. Most specifically, label the information given about any angles in the picture. Also, the question states that the quadrilateral is cyclic, so why not sketch a circle around it as a reminder?

    In this question the angles B^ and C^ are given.


    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    The key to this question is the fact that the quadrilateral is cyclic: that means that opposite angles are supplementary. We are trying to find D^, and it must be related to B^ like this: D^+B^=180.

    Substitute in the expression given for B^ and solve for the angle you want: D^+(x35)=180. This works out to be: D^=180(x35)=x+215.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As explained in the solution above, the answer to the question comes from the theorem about opposite angles in cyclic quadrilaterals. The correct choice is, 'Opposite angles of a cyclic quadrilateral.'


    Submit your answer as:

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment RW¯ touching the circle at point P. There are four points on the circle: P,Q,S and T, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles is congruent to angle χ? If there are no angles congruent to χ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

The figure below shows the tangent line in red, as well as the angle we are dealing with in the question, angle χ.

In this case, angle χ is not between the tangent line and a chord (it is between two of the chords - see above). The tangent-chord theorem does not tell us anything about this angle: it is not connected to a specific angle in the figure.

While there are other angles in the figure which are congruent to each other, we do not know anything useful about angle χ: the correct answer is 'None of the above.'


Submit your answer as:

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, PQRT is a cyclic quadrilateral having RTPQ. The tangent at P meets RT produced at S. PQ=QT and PT^S=69°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why P^2=69°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that PT is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines RT and PQ. The line PT is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, P^2 and T^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    P^2=T^1

    T^1 is given in the question as 69°.

    P^2=T^1P2=69°

    P^2 must be equal to 69° because alternate interior angles are equal. The correct answer is Option D.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. Q^1
    2. P^1
    Answer:
    1. Q^1= °
      Reason:
    2. P^1= °
      Reason:
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find Q^1 using the fact that triangle PQT is isosceles.


    STEP: Find Q^1 using the fact that triangle PQT is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that P^2=69°. We also know from the question that triangle PQT is isosceles with P^2 equal to T^2. So that means T^2 must also be 69°.

    Now we know two of the angles in triangle PQT so we can calculate the size of Q^1:

    Q^1=180°2(69°)=42°

    To find both angles in triangle PQT, we used the sum of angles in a triangle.


    STEP: Find P^1 using the tangent chord theorem
    [−2 points ⇒ 0 / 5 points left]

    The fact that PS is a tangent line at Point P means we can consider theorems about tangent lines. According to the tangent chord theorem, the angle between a tangent and a chord (in this case, the chord is PT) is equal to an angle on the circle's circumference subtended by that chord. In this diagram, that means P^1 is equal to Q^1:

    P^1=Q^1

    We know from our work above that Q^1=42°. So:

    P^1=42°

    The key reason for this information is the tangent chord theorem.


    Submit your answer as: andandand

Using the tangent-chord theorem to find angles

Line segment BE¯ is tangent to circle O at point D. Points A and G lie on the circle. Two angles are given: ED^G=43° and BD^A=68°.

Determine the values of the three angles in the triangle, a,b and c.

Answer:

a=°
b=°
c=°

numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle a, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle a
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: a makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

43°+a+68°=180°111°+a=180°a=69°

Cool: we have one of the three answers: a=69°.


STEP: Connect information using the tangent-chord theorem to find c
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both b and c to angles that we know. For example, let's look at the 43° angle : it reaches from the tangent line to chord DG¯. From the chord, we find angle c across the triangle. This is the angle which must be congruent to the 43° angle. (See the green parts of the figure below.) So c=43°.


STEP: Repeat the previous step to find b
[−1 point ⇒ 0 / 3 points left]

The 68° angle and b are related in exactly the same way. The diagram shows the connection between these angles in purple. So b=68°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: a=69°,b=68° and c=43°.


Submit your answer as: andand

Two tangents from a single point

In the figure below, two line segments run from point I to a circle. Both segments are tangent to the circle: the tangent points are G and H. The centre of the circle is at point O and the radius of the circle is 17,9 cm. Line segment HI¯ is 31 cm long. (The figure is drawn to scale!)

Find the length of GI¯ and select what type of angle IG^O is (right angle, obtuse, acute, etc.).

Answer: GI¯= cm
IG^O is
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point I must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, GI¯HI¯, which means that GI¯=31 cm.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle IG^O, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point G.

Is IG^O larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

In this case, segment GO¯ is a radius which touches the tangent point! That means that IG^O must be a right angle (it has a measure of 90 degrees).


Submit your answer as: and

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Three points lie on the circle: D,C and A. There are also two points shown outside of the circle. There are two lines tangent to the circle: they touch the circle at points D and C.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    You can see there are two different points outside of the circle: B and E. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

    In this case, the diagram does include a pair of tangent lines: they pass from point B and are tangent to the circle at points D and C. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: BC;BD.


    Submit your answer as:
  2. Name one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? The red line segment shown starts at the centre of the circle, but at the spot where it touches the circle, there is no tangent line.

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

Exercises

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line BE¯ which touches the circle at point D. Three points lie on the circle, D,A and G, and five angles are labelled: α,β,θ, etc.

Idenitify an angle which is congruent to angle β. (If there is no angle congruent to β, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (BE¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, D: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord - in this question it is angle β
  4. another angle subtending the chord

In the figure below, the relevant information is shown in red. It is very helpful to see the figure below as a tangent line and a triangle - the tangent-chord theorem will always include that triangle!

Notice that one of the angles is inside the triangle, while the other angle is outside of the triangle.

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle β to the red chord... then from the red chord follow the red arrow across the circle to angle θ which subtends the chord (angle θ is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles we just connected are congruent (the same size): angle β angle θ. By looking at these two angles, you can see that they do look like they are about the same size; the tangent-chord theorem says that they must be the same size!


Submit your answer as:

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line RW¯ which touches the circle at point P. Three points lie on the circle, P,S and T, and five angles are labelled: 1,2,3, etc.

Idenitify an angle which is congruent to angle 3. (If there is no angle congruent to 3, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (RW¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, P: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

The figure below shows the tangent line in red, as well as the angle we are dealing with in the question, angle 3.

In this case, angle 3 is not between the tangent line and a chord (it is between the two chords - see above). The tangent-chord theorem does not tell us anything about this angle: it is not connected to a specific angle in the figure.

While there are other angles in the figure which are congruent to each other, we do not know anything useful about angle 3: the correct answer is 'None of the above.'


Submit your answer as:

The tangent-chord theorem: finding congruent angles

The diagram below is drawn to scale. It shows a circle with its centre at O and a tangent line IM¯ which touches the circle at point G. Three points lie on the circle, G,J and K, and five angles are labelled: 1,2,3, etc.

Idenitify an angle which is congruent to angle 2. (If there is no angle congruent to 2, choose 'None of the above'.)

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
When you use the tangent-chord theorem, you must first identify the tangent line and the chord you should use. The tangent line (IM¯) is not too hard to spot. To find the correct chord, put your finger on the tangent point, G: you can see there are two chords from that point. You can use the tangent-chord theorem if the angle you know or the angle you want sits between the tangent line and one of the chords.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question uses the tangent-chord theorem, which connects four things:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord - in this question it is angle 2
  4. another angle subtending the chord

In the figure below, the relevant information is shown in red. It is very helpful to see the figure below as a tangent line and a triangle - the tangent-chord theorem will always include that triangle!

Notice that one of the angles is inside the triangle, while the other angle is outside of the triangle.

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle 2 to the red chord... then from the red chord follow the red arrow across the circle to angle 1 which subtends the chord (angle 1 is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles we just connected are congruent (the same size): angle 2 angle 1. By looking at these two angles, you can see that they do look like they are about the same size; the tangent-chord theorem says that they must be the same size!


Submit your answer as:

Applying theorems from Euclidean geometry

The figure below shows circle O with two tangent lines touching the circle at points G and H. The tangent lines meet at point I. HI¯ has a length of c+55 and the radius of the circle is 2c+35.

  1. Find the length of GI¯ in terms of c. If the answer cannot be determined, write 'No solution'.

    Answer: length of GI¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    When you read "in terms of c," it means that the answer will be a maths expression which includes the variable c.
    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is 2c+35, as well as the length of the line segment HI¯=c+55. Put this information on the diagram - when you label the diagram it helps to organise the information.

    There are also some congruent (equal) line segments in the figure: you should mark them to show that the segments are equal in length. One pair of equal segments are the radii (see below).


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now think about what the question is asking you to find: you must determine the length of GI¯. What comparisons can you make to determine the length of that segment? There is a theorem about tangent lines with circles which might be helpful: if two tangent lines to a circle come from the same point, then they have equal lengths.

    There are two tangent lines in this figure, and they come from the same place (point I). Therefore the segments are congruent: GI¯HI¯ (the symbol means 'congruent'). Therefore, the answer is GI¯=c+55.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As described above, the correct answer is: Two tangents drawn from the same point outside a circle.


    Submit your answer as:

Applying theorems from Euclidean geometry

The figure below shows a circle with centre at point O. There are two points on the circle, V and W, and the line from X to W is a tangent line to the circle. VX¯ has a length of y220 and the radius of the circle is 3y475.

  1. Determine the diameter of the circle in terms of y. If the answer cannot be determined, write 'No solution'.

    Answer: The diameter is:
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    When you read "in terms of y," it means that the answer will be a maths expression which includes the variable y.
    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is 3y475, as well as the length of the line segment VX¯=y220. Put this information on the diagram - when you label the diagram it helps to organise the information.


    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    Now think about what the question is asking you to find: the question asks for the length of the diameter. The diameter is always twice as long as the radius: d=2r.

    To determine the answer, you just need to take the radius, which is given, and double it: d=2(3y475)=3y2150.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    However, this question does not use any special theorem: you just use the fact that the diameter is twice as long as the radius. The correct answer is: 'None of the above.'


    Submit your answer as:

Applying theorems from Euclidean geometry

The figure below shows circle O with two tangent lines touching the circle at points V and W. The tangent lines meet at point X. VX¯ has a length of 3y2+80 and the radius of the circle is 4y3+35.

  1. Find the length of WX¯ in terms of y. If the answer cannot be determined, write 'No solution'.

    Answer: length of WX¯=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    When you read "in terms of y," it means that the answer will be a maths expression which includes the variable y.
    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    A good start for any geometry question is to add all information you know about the shapes in the figure: are there any things you can label in this figure? Indeed, the question tells you the radius of the circle is 4y3+35, as well as the length of the line segment VX¯=3y2+80. Put this information on the diagram - when you label the diagram it helps to organise the information.

    There are also some congruent (equal) line segments in the figure: you should mark them to show that the segments are equal in length. One pair of equal segments are the radii (see below).


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now think about what the question is asking you to find: you must determine the length of WX¯. What comparisons can you make to determine the length of that segment? There is a theorem about tangent lines with circles which might be helpful: if two tangent lines to a circle come from the same point, then they have equal lengths.

    There are two tangent lines in this figure, and they come from the same place (point X). Therefore the segments are congruent: WX¯VX¯ (the symbol means 'congruent'). Therefore, the answer is WX¯=3y2+80.


    Submit your answer as:
  2. Which of the following theorems did you use to answer the question above?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    Check the solution above to figure out which theorem you used... or if you did not use any of the theorems on this list.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In maths, and in Euclidean geometry in particular, it is important to know the reason that allows you to find some piece of information. In this question, if the reason is one of the theorems you need to know which one it is!

    As described above, the correct answer is: Two tangents drawn from the same point outside a circle.


    Submit your answer as:

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment UX¯ touching the circle at point V. There are four points on the circle: V,W,Y and Z, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles is congruent to angle κ? If there are no angles congruent to κ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle κ

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect these pieces, move through the figure like this: from the tangent line, rotate through angle θ and angle δ up to chord VY¯ - you need to ignore VW¯! Then from the red chord go across the triangle to angle κ which subtends the chord (angle κ is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): κ (δ+θ).


Submit your answer as:

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment IL¯ touching the circle at point K. There are four points on the circle: K,J,H and G, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles is congruent to angle χ? If there are no angles congruent to χ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle χ

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle β to the red chord... then from the red chord follow the red arrow across the circle to angle χ which subtends the chord (angle χ is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): χ β.


Submit your answer as:

The tangent-chord theorem: messy situations

Below is circle O with a tangent line segment BE¯ touching the circle at point D. There are four points on the circle: D,C,A and G, and the following angles are labelled: α,β,δ, etc. (The diagram is drawn to scale.)

Which of the angles is congruent to angle χ? If there are no angles congruent to χ, choose 'None of the above'.

Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Notice that the circle has two triangles in it: try to see these two triangles separately


STEP: Identify parts of the figure related to the tangent-chord theorem
[−1 point ⇒ 0 / 1 points left]

This question is about the tangent-chord theorem, which connects four things:

  1. a tangent line for a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord - in this question it is angle χ

Even in a messy diagram like this one, these four things will sit within the tangent line and a triangle.

In the figure below, the relevant information is shown in red, and the important triangle is shaded. (The other triangle is useless in this question!)

To connect all of these pieces, move through the figure like this: from the tangent line, follow the red arc of angle θ to the red chord... then from the red chord follow the red arrow across the circle to angle χ which subtends the chord (angle χ is circled). These are the pieces related by the tangent-chord theorem.

The tangent-chord theorem tells us that the two angles connected as above must be congruent (the same size): χ θ.


Submit your answer as:

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, PQRT is a cyclic quadrilateral having RTPQ. The tangent at P meets RT produced at S. PQ=QT and PT^S=70°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why P^2=70°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that PT is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines RT and PQ. The line PT is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, P^2 and T^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    P^2=T^1

    T^1 is given in the question as 70°.

    P^2=T^1P2=70°

    P^2 must be equal to 70° because alternate interior angles are equal. The correct answer is Option A.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. Q^1
    2. T^3
    Answer:
    1. Q^1= °
      Reason:
    2. T^3= °
      Reason:
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find Q^1 using the fact that triangle PQT is isosceles.


    STEP: Find Q^1 using the fact that triangle PQT is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that P^2=70°. We also know from the question that triangle PQT is isosceles with P^2 equal to T^2. So that means T^2 must also be 70°.

    Now we know two of the angles in triangle PQT so we can calculate the size of Q^1:

    Q^1=180°2(70°)=40°

    To find both angles in triangle PQT, we used the sum of angles in a triangle.


    STEP: Find T^3 using angles on a straight line
    [−2 points ⇒ 0 / 5 points left]

    At Point T there are three angles, and they make a straight line. So they must have a sum of 180°.

    180°=T^1+T^2+T^3

    We know from the diagram that T^1=70°. And from the work above we know that T^2 is also equal to 70°.

    180°=(70°)+(70°)+T^340°=T^3
    NOTE: We can also get this answer using alternate interior angles (on parallel lines) with Q^1. But even though that is an accurate reason, it is not one of the answer options.

    The key reason for this information is angles on a straight line.


    Submit your answer as: andandand

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, PQRT is a cyclic quadrilateral having PQRT. The tangent at R meets PQ produced at S. RT=QT and RQ^S=69°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why R^1=69°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that QR is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines PQ and RT. The line QR is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, R^1 and Q^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    R^1=Q^1

    Q^1 is given in the question as 69°.

    R^1=Q^1R1=69°

    R^1 must be equal to 69° because alternate interior angles are equal. The correct answer is Option C.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. T^1
    2. R^2
    Answer:
    1. T^1= °
      Reason:
    2. R^2= °
      Reason:
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find T^1 using the fact that triangle QRT is isosceles.


    STEP: Find T^1 using the fact that triangle QRT is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that R^1=69°. We also know from the question that triangle QRT is isosceles with R^1 equal to Q^2. So that means Q^2 must also be 69°.

    Now we know two of the angles in triangle QRT so we can calculate the size of T^1:

    T^1=180°2(69°)=42°

    To find both angles in triangle QRT, we used the sum of angles in a triangle.


    STEP: Find R^2 using the tangent chord theorem
    [−2 points ⇒ 0 / 5 points left]

    The fact that RS is a tangent line at Point R means we can consider theorems about tangent lines. According to the tangent chord theorem, the angle between a tangent and a chord (in this case, the chord is QR) is equal to an angle on the circle's circumference subtended by that chord. In this diagram, that means R^2 is equal to T^1:

    R^2=T^1

    We know from our work above that T^1=42°. So:

    R^2=42°

    The key reason for this information is the tangent chord theorem.


    Submit your answer as: andandand

Triangles and circles

Adapted from DBE Nov 2016 Grade 12, P2, Q8.1
Maths formulas

In the diagram below, ABCE is a cyclic quadrilateral having CEAB. The tangent at A meets CE produced at D. AB=BE and AE^D=72°.

NOTE: Do not assume that the diagram is drawn to scale.
  1. Give a reason why A^1=72°. Select the most appropriate answer from the choices below.

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the parallel lines in the figure. Notice that AE is a transversal.


    STEP: Use the parallel lines
    [−1 point ⇒ 0 / 1 points left]

    The figure includes the parallel lines CE and AB. The line AE is a transversal crossing these parallel lines. So we can use what we know about transversal geometry. This might include corresponding angles, alternate interior angles, and so on.

    In this case, A^1 and E^1 are alternate interior angles. And alternate interior angles on parallel lines are equal:

    A^1=E^1

    E^1 is given in the question as 72°.

    A^1=E^1A1=72°

    A^1 must be equal to 72° because alternate interior angles are equal. The correct answer is Option B.


    Submit your answer as:
  2. Calculate each of the following values. For each value, identify the most significant reason for that value from the list of choices.

    1. B^1
    2. E^3
    Answer:
    1. B^1= °
      Reason:
    2. E^3= °
      Reason:
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    You can find B^1 using the fact that triangle ABE is isosceles.


    STEP: Find B^1 using the fact that triangle ABE is isosceles
    [−3 points ⇒ 2 / 5 points left]

    From Question 1 we know that A^1=72°. We also know from the question that triangle ABE is isosceles with A^1 equal to E^2. So that means E^2 must also be 72°.

    Now we know two of the angles in triangle ABE so we can calculate the size of B^1:

    B^1=180°2(72°)=36°

    To find both angles in triangle ABE, we used the sum of angles in a triangle.


    STEP: Find E^3 using angles on a straight line
    [−2 points ⇒ 0 / 5 points left]

    At Point E there are three angles, and they make a straight line. So they must have a sum of 180°.

    180°=E^1+E^2+E^3

    We know from the diagram that E^1=72°. And from the work above we know that E^2 is also equal to 72°.

    180°=(72°)+(72°)+E^336°=E^3
    NOTE: We can also get this answer using alternate interior angles (on parallel lines) with B^1. But even though that is an accurate reason, it is not one of the answer options.

    The key reason for this information is angles on a straight line.


    Submit your answer as: andandand

Using the tangent-chord theorem to find angles

Line segment UX¯ is tangent to circle O at point V. Points Y and Z lie on the circle. Two angles are given: UV^Z=50° and XV^Y=50°.

Find the values of the three angles in the triangle, x,y and z.

Answer:

x=°
y=°
z=°

numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle y, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle y
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: y makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

50°+y+50°=180°100°+y=180°y=80°

Cool: we have one of the three answers: y=80°.


STEP: Connect information using the tangent-chord theorem to find z
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both x and z to angles that we know. For example, let's look at the 50° angle on the left of the tangent point: it reaches from the tangent line to chord VZ¯. From the chord, we find angle z across the triangle. This is the angle which must be congruent to the 50° angle. (See the green parts of the figure below.) So z=50°.


STEP: Repeat the previous step to find x
[−1 point ⇒ 0 / 3 points left]

The other 50° angle and x are related in exactly the same way. The diagram shows the connection between these angles in purple. So x=50°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: x=50°,y=80° and z=50°.


Submit your answer as: andand

Using the tangent-chord theorem to find angles

Line segment IL¯ is tangent to circle O at point K. Points H and G lie on the circle. Two angles are given: LK^G=52° and IK^H=52°.

Find the values of the three angles in the triangle, a,b and c.

Answer:

a=°
b=°
c=°

numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle b, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle b
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: b makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

52°+b+52°=180°104°+b=180°b=76°

Cool: we have one of the three answers: b=76°.


STEP: Connect information using the tangent-chord theorem to find c
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both a and c to angles that we know. For example, let's look at the 52° angle on the left of the tangent point: it reaches from the tangent line to chord KG¯. From the chord, we find angle c across the triangle. This is the angle which must be congruent to the 52° angle. (See the green parts of the figure below.) So c=52°.


STEP: Repeat the previous step to find a
[−1 point ⇒ 0 / 3 points left]

The other 52° angle and a are related in exactly the same way. The diagram shows the connection between these angles in purple. So a=52°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: a=52°,b=76° and c=52°.


Submit your answer as: andand

Using the tangent-chord theorem to find angles

Line segment BE¯ is tangent to circle O at point D. Points A and G lie on the circle. Two angles are given: ED^G=46° and BD^A=46°.

Determine the values of the three angles in the triangle, α,β and θ.

Answer:

α=°
β=°
θ=°

numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the size of angle α, which makes a straight line with the two angles on each side of it.


STEP: Determine the size of angle α
[−1 point ⇒ 2 / 3 points left]

We can solve this question using the tangent-chord theorem. However, there is one angle we can get very quicky: α makes a straight line (a straight angle) with the two given angles, so they must have a sum of 180°. Write an equation to summarize this, and then solve it.

46°+α+46°=180°92°+α=180°α=88°

Cool: we have one of the three answers: α=88°.


STEP: Connect information using the tangent-chord theorem to find θ
[−1 point ⇒ 1 / 3 points left]

To find the other angles, we need the tangent-chord theorem. This theorem ties four things together:

  1. a tangent line to a circle
  2. a chord touching the tangent point
  3. the angle between the tangent line and the chord
  4. another angle subtending the chord

All of these things are sitting from the tangent line over and into the triangle in the figure.

In this case, we can use the theorem twice: we can link both β and θ to angles that we know. For example, let's look at the 46° angle on the left of the tangent point: it reaches from the tangent line to chord DG¯. From the chord, we find angle θ across the triangle. This is the angle which must be congruent to the 46° angle. (See the green parts of the figure below.) So θ=46°.


STEP: Repeat the previous step to find β
[−1 point ⇒ 0 / 3 points left]

The other 46° angle and β are related in exactly the same way. The diagram shows the connection between these angles in purple. So β=46°.

That is the end - we have all three angles. Notice that these angles, which are inside of a triangle, have a sum of 180°. That must be true because they are in a triangle... and that is why maths is awesome!

The three answers are: α=88°,β=46° and θ=46°.


Submit your answer as: andand

Two tangents from a single point

In the figure below, two line segments stretch from point X to a circle. Both segments are tangent to the circle: the tangent points are V and W. The centre of the circle is at point O and the radius of the circle is 11 m. Line segment VX¯ is 25,2 m long. The length of VY¯ is 12,1 m. (The figure is drawn to scale!)

Determine the length of WX¯ and select what type of angle XV^Y is (right angle, obtuse, acute, etc.).

Answer: WX¯= m
XV^Y is
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point X must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, WX¯VX¯, which means that WX¯=25,2 m.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle XV^Y, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point V.

Is XV^Y larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

The angle XV^Y is smaller than the right angle: the correct answer is 'acute.'


Submit your answer as: and

Two tangents from a single point

In the figure below, two line segments stretch from point I to a circle. Both segments are tangent to the circle: the tangent points are K and J. The centre of the circle is at point O and the radius of the circle is 15 cm. Line segment IK¯ is 24,2 cm long. The length of HK¯ is 26,3 cm. (The figure is drawn to scale!)

Determine the length of IJ¯ and select what type of angle IK^H is (right angle, obtuse, acute, etc.).

Answer: IJ¯= cm
IK^H is
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point I must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, IJ¯IK¯, which means that IJ¯=24,2 cm.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle IK^H, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point K.

Is IK^H larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

You can see in the figure that IK^H is larger than the right angle. Therefore, the angle is obtuse.


Submit your answer as: and

Two tangents from a single point

In the figure below, two line segments stretch from point I to a circle. Both segments are tangent to the circle: the tangent points are G and H. The centre of the circle is at point O and the radius of the circle is 3 m. Line segment HI¯ is 5,2 m long. The length of GJ¯ is 3,7 m. (The figure is drawn to scale!)

Find the length of GI¯ and select what type of angle IG^J is (right angle, obtuse, acute, etc.).

Answer: GI¯= m
IG^J is
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The most important thing to notice in this question is that there are two tangent lines which come from the same point: we have a theorem about this situation! Two tangents to a circle from a single point must be the same size. Thus, the two tangents from point I must be congruent (the same length) to each other. This is shown below with the purple lines, and the '||' markers on the line segments.

Therefore, GI¯HI¯, which means that GI¯=5,2 m.


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The second question is about the angle IG^J, which is shown in the diagram above by the red arc.

You must remember that a tangent line for a circle is always perpendicular to the radius of the circle which touches the tangent point. This right angle is shown in red, where you can also see a radius drawn in to point G.

Is IG^J larger or smaller than the right angle? Obtuse angles are greater than 90 and acute angles are less than 90 degrees.

You can see in the figure that IG^J is larger than the right angle. Therefore, the angle is obtuse.


Submit your answer as: and

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Three points lie on the circle: K,J and H. There are also two points shown outside of the circle. There are two lines tangent to the circle: they touch the circle at points K and J.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    You can see there are two different points outside of the circle: I and L. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

    In this case, the diagram does include a pair of tangent lines: they pass from point I and are tangent to the circle at points K and J. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: IJ;IK.


    Submit your answer as:
  2. Identify one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? The red line segment shown starts at the centre of the circle, but at the spot where it touches the circle, there is no tangent line.

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Three points lie on the circle: P,Q and S. There are also two points shown outside of the circle. There are two lines tangent to the circle: they touch the circle at points P and Q.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    You can see there are two different points outside of the circle: R and W. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

    In this case, the diagram does include a pair of tangent lines: they pass from point R and are tangent to the circle at points P and Q. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: QR;PR.


    Submit your answer as:
  2. Name one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? The red line segment shown starts at the centre of the circle, but at the spot where it touches the circle, there is no tangent line.

    NOTE: It is possible that there are some right angles in the figure. However, we cannot know that they are perpendicular without a reason: we need some kind of proof! Without a reason (like a theorem or a calculation) we cannot be sure.

    Based on what you can see in the diagram, the answer is none.


    Submit your answer as:

What do congruent tangent lines look like?

The diagram below is drawn to scale. It shows a circle with its centre at O. Three points lie on the circle: D,C and A. There are also two points shown outside of the circle. There are two lines tangent to the circle: they touch the circle at points D and C.

  1. Name two congruent line segments (segments with equal lengths) in the figure.

    INSTRUCTION: Name the segments like this: AB ; CD. If there are no congruent line segments, write none.
    Answer: Two congruent segments are: .
    list
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Sometimes geometry figures can be very busy! It is a good idea to look at some parts of the pictures separately from the rest. You may even want to use your fingers to cover up some lines: this will help to simplify the picture. Move your hand around different places and see if it helps.


    STEP: Look for two congruent tangent line segments
    [−2 points ⇒ 0 / 2 points left]

    The word 'congruent' means identical: same size, shape, etc. To find two congruent line segments, we need to find two line segments with the same length.

    One of the circle theorems says that if there are two tangent lines to a circle from the same point, then they must be the same length. Are there two tangent lines like that in this figure?

    To test if there are two equal tangent lines, start at a point outside the circle: then look to see if there are two lines from that point stretching toward the circle... and if they are both tangent to the circle. If there are not two lines, or if the are not both tangent lines, then we cannot know that there are equal segments in the diagram.

    You can see there are two different points outside of the circle: B and E. Check each of them to see if there are two lines reaching the circle, and if those lines are tangent to the circle.

    In this case, the diagram does include a pair of tangent lines: they pass from point B and are tangent to the circle at points D and C. You can see these segments shown in red in the diagram above. They are also marked as congruent.

    NOTE: There may be other line segments in the diagram which look like they are the same size - in fact they might even be congruent! However, we cannot know that they are congruent without a reason, like the theorem about tangents from a single point. Without a reason (like a theorem) we must leave them.

    The correct answer is: BC;BD.


    Submit your answer as:
  2. Name one 90° angle in the figure (as it is shown).

    INSTRUCTION: Name the angle by three points, for example, AOB. If there are no 90° angles in the figure, write none.
    Answer:
    stringset
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Remember that a tangent line is always perpendicular to the radius reaching the tangent point... In this figure, is there a radius in the circle which touches a tangent point? If so, then there is a right angle there!


    STEP: Look for a right angle at a point of tangency
    [−1 point ⇒ 0 / 1 points left]

    One thing you must remember about tangent lines for circles is this fact: the tangent line will be perpendicular to the radius touching the tangent point.

    Any radius must come from the centre of the circle. Are there any lines starting at point O in this figure? Yes, there is a radius which goes to point C; see the red radius below.

    Based on what you can see in the diagram, the answer is BC^O; this can also be written as OC^B.


    Submit your answer as:

5. Mixed applications

Exercises